The `C_("rms")` of He `5.3 xx 10^(2) ms^(-1)` and the distance per collision is `6.18 xx 10^(-8)` metre `"collision"^(-1)`. The collision frequency is `x xx 10^(9)` collision `sec^(-1)`. What will be the value of x?

Answer

To find the collision frequency (ν) of helium gas, we can use the formula:

\[ \nu = \frac{C_{rms}}{d} \]

where:
- \(C_{rms}\) is the root mean square speed,
- \(d\) is the distance per collision.

Given:
- \(C_{rms} = 5.3 \times 10^2 \, \text{m/s}\)
- \(d = 6.18 \times 10^{-8} \, \text{m}\)

### Step 1: Substitute the values into the formula
We can substitute the given values into the formula for collision frequency:

\[ \nu = \frac{5.3 \times 10^2 \, \text{m/s}}{6.18 \times 10^{-8} \, \text{m}} \]

### Step 2: Perform the division
Now, we will perform the division:

\[ \nu = \frac{5.3}{6.18} \times \frac{10^2}{10^{-8}} \]

Calculating the numerical part:

\[ \frac{5.3}{6.18} \approx 0.857 \]

Calculating the powers of ten:

\[ 10^2 \div 10^{-8} = 10^{2 - (-8)} = 10^{2 + 8} = 10^{10} \]

### Step 3: Combine the results
Now, we can combine the results:

\[ \nu \approx 0.857 \times 10^{10} \, \text{collisions/sec} \]

### Step 4: Express in the required format
We need to express the frequency in the form \(x \times 10^9\):

\[ \nu \approx 8.57 \times 10^9 \, \text{collisions/sec} \]

Thus, the value of \(x\) is approximately:

\[ x \approx 8.57 \]

### Final Answer:
The value of \(x\) is approximately \(8.57\).

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